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Question #42783

Difficulty: 8/10 | 🟢 Free

The Question

An incompressible, inviscid, perfectly conducting liquid of mass density $\rho$ flows steadily along the grounded, horizontal bottom plate of a parallel-plate capacitor. The top plate is at a height $d$ and is maintained at a constant positive potential $V$ by a battery. The space between the liquid surface and the top plate is a vacuum. At the inlet section, the liquid has a uniform depth $h_1$ and velocity $v_1$ . At a downstream section, the depth is $h_2$ . Assuming the flow is uniform across the depth, the electric field is purely vertical, and the liquid remains in electrical contact with the bottom plate, which of the following expressions gives the square of the velocity, $v_2^2$ , at the downstream section? (Let $g$ be the acceleration due to gravity and $\epsilon_0$ be the vacuum permittivity).

Options

A.
$v_1^2 + 2g(h_1 - h_2) + \frac{\epsilon_0 V^2}{\rho} \left[ \frac{1}{(d - h_2)^2} - \frac{1}{(d - h_1)^2} \right]$
B.
$v_1^2 + 2g(h_1 - h_2) - \frac{\epsilon_0 V^2}{\rho} \left[ \frac{1}{(d - h_2)^2} - \frac{1}{(d - h_1)^2} \right]$
C.
$v_1^2 + 2g(h_1 - h_2)$
D.
$v_1^2 + 2g(h_1 - h_2) + \frac{2\epsilon_0 V^2}{\rho} \left[ \frac{1}{(d - h_2)^2} - \frac{1}{(d - h_1)^2} \right]$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Common Potential Redistribution|Equation of Continuity|Bernoulli’s Energy Balance

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of fluid dynamics (Bernoulli's Principle) with electrostatic energy density at a moving boundary.

The primary trap lies in misidentifying the boundary pressure.

Students often incorrectly assume the pressure at the liquid surface is zero (vacuum) and ignore the electrostatic traction.

Another common mistake is a sign error, where students assume the electric field pushes the liquid down rather than pulling the induced surface charges upward.

Furthermore, if a student incorrectly models the capacitor as an isolated system (constant charge) instead of a constant-potential system maintained by a battery, they would erroneously conclude that the electric field $E$ is independent of $h$ , causing the electrostatic pressure terms to cancel out entirely.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Common Potential Redistribution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free